Differential of norm with symmetric matrix

Question

Let $z: I \rightarrow \mathbb{R}^n$ be a differentiable function. We know that $$\frac{d}{dt} |z(t)|^2= 2\langle z'(t),z(t) \rangle.$$ Now, let $A$ be any $n\times n$ matrix (maybe symmetric, definite positive...etc). Do we have something similar such that the differential is $2\langle z'(t),Az(t) \rangle$? Any remark would be helpful (equality, inequality...etc).

Answer 1

Let $I \subset \Bbb{R}$ be an open set, and let $A \in M_{n \times n}(\Bbb{R})$ be a symmetric matrix. If you assume $z:I \to \Bbb{R}^n$ is differentiable, then you can show that the function $f: I \to \Bbb{R}$ defined by \begin{align} f(t) = \langle A z(t) , z(t)\rangle \end{align} is differentiable and by the chain rule, we have that for every $t \in I$, \begin{align} f'(t) &= \langle A z'(t), z(t) \rangle + \langle A z(t) , z'(t)\rangle \\ &= 2 \langle A z(t) , z'(t)\rangle, \end{align} the last equality is because $A$ is symmetric.

Answer 2

Suppose

$y(t) = \langle z(t), Az(t) \rangle; \tag 1$

then

$\dot y(t) = \langle \dot z(t), Az(t) \rangle + \langle z(t), A\dot z(t) \rangle$ $ = \langle \dot z(t), Az(t) \rangle + \langle A^Tz(t), \dot z(t) \rangle = \langle \dot z(t), Az(t) \rangle + \dot z(t), A^T z(t) \rangle$ $= \langle \dot z(t), (A + A^T)z(t) \rangle; \tag 2$

when $A$ is symmetric,

$A = A^T, \tag 3$

this becomes

$\dot y(t) = 2\langle \dot z(t), Az(t) \rangle. \tag 4$

Nota Bene: It is also worth observing that the operator $A$ occurring in (2), (4) may be transferred to $\dot z(t)$ via the usual properties of the matrix transpose, viz.

$\dot y(t) = \langle (A + A^T)\dot z(t), z(t) \rangle = \langle z(t), (A + A^T)\dot z(t) \rangle, \tag 5$

and

$\dot y(t) = 2\langle A\dot z(t), z(t) \rangle = 2 \langle z(t), A\dot z(t) \rangle \tag 4$

in the event that $A = A^T$ is symmetric. End of Note.